Motion in Gravitational Fields

Newton's law of gravitation gives $F = \dfrac{GMm}{r^2}$, so $F \propto \dfrac{1}{r^2}$. Tripling $r$ divides the force by $3^2 = 9$.

The gravitational field $g$ at any point is the force per unit mass on a test mass, which always points toward the source. So the field lines point radially inward, getting weaker with distance ($g \propto 1/r^2$).

At the station's altitude gravity is still about 90% of its surface value. The astronauts feel weightless because they and the station are continuously falling toward Earth together (in free fall), so there is no normal force between them. It is "apparent" weightlessness, not an absence of gravity.

$F = \dfrac{GMm}{r^2} = \dfrac{6.67\times10^{-11} \times 6.0\times10^{24} \times 100}{(1.0\times10^{7})^2} = \dfrac{4.0\times10^{16}}{1.0\times10^{14}} = 400\text{ N}$.

$g = \dfrac{GM}{r^2} = \dfrac{6.67\times10^{-11} \times 6.4\times10^{23}}{(3.4\times10^{6})^2} = \dfrac{4.27\times10^{13}}{1.16\times10^{13}} \approx 3.7\text{ N/kg}$ (about $0.38$ of Earth's).

Since $g \propto \dfrac{1}{r^2}$, doubling the distance from the centre divides the field by $2^2 = 4$: $g = \dfrac{9.8}{4} = 2.45\text{ N/kg}$.

Gravity provides the centripetal force, giving $v = \sqrt{\dfrac{GM}{r}} = \sqrt{\dfrac{4.0\times10^{14}}{8.0\times10^{6}}} = \sqrt{5.0\times10^{7}} \approx 7.1\times10^{3}\text{ m/s}$. The orbital speed does not depend on the satellite's mass.

$T = \dfrac{2\pi r}{v} = \dfrac{2\pi (8.0\times10^{6})}{7.1\times10^{3}} = \dfrac{5.03\times10^{7}}{7.1\times10^{3}} \approx 7.1\times10^{3}\text{ s}$ (about two hours).

$\dfrac{T_2}{T_1} = \left(\dfrac{r_2}{r_1}\right)^{3/2} = 4^{3/2} = 8$, so the outer satellite has a period of $8T$. A larger orbit means a longer path and a slower speed.

Kepler's third law rearranged gives $M = \dfrac{4\pi^2 r^3}{GT^2} = \dfrac{4\pi^2 (2.0\times10^{8})^3}{6.67\times10^{-11}(1.5\times10^{5})^2} = \dfrac{3.16\times10^{26}}{1.50} \approx 2.1\times10^{26}\text{ kg}$.

$v_{esc} = \sqrt{\dfrac{2GM}{r}} = \sqrt{2}\,\sqrt{\dfrac{GM}{r}} = \sqrt{2}\,v_{orbit}$. So $v_{esc} = 1.41 \times 7.1\times10^{3} \approx 1.0\times10^{4}\text{ m/s}$. Escape speed is always $\sqrt{2}$ times the circular orbital speed.

Gravitational potential energy is $U = -\dfrac{GMm}{r}$, negative and growing toward zero as $r$ increases. Moving to a higher orbit (larger $r$) makes $U$ less negative, i.e. it increases; work must be done against gravity.

The total energy is $E = KE + U = \dfrac{GMm}{2r} - \dfrac{GMm}{r} = -\dfrac{GMm}{2r}$, which is negative. A negative total energy is the signature of a bound orbit; a body with zero or positive total energy would escape.

In a circular orbit $KE = \dfrac{GMm}{2r}$ while $U = -\dfrac{GMm}{r}$. So $KE = -\tfrac{1}{2}U$: the kinetic energy is half the magnitude of the (negative) potential energy. It also equals $-E$, the magnitude of the total energy.

Orbital speed is $v = \sqrt{\dfrac{GM}{r}}$, which decreases as $r$ increases. So the lower satellite (smaller $r$) moves faster. Low-orbit satellites race around Earth, while distant ones drift slowly.

A geostationary satellite stays fixed above one point on the equator, so its period must match Earth's rotation: $24$ hours. This requires a specific high orbit (about $36{,}000\text{ km}$ altitude), far above a low-orbit satellite that circles in about 90 minutes.

From Kepler, $r = \left(\dfrac{GM\,T^2}{4\pi^2}\right)^{1/3} = \left(\dfrac{4.0\times10^{14}(86400)^2}{4\pi^2}\right)^{1/3} \approx (7.6\times10^{22})^{1/3} \approx 4.2\times10^{7}\text{ m}$ (around $36{,}000\text{ km}$ above the surface).

Orbital speed is $v = \sqrt{\dfrac{GM}{r}} \propto \dfrac{1}{\sqrt{r}}$. Increasing the radius by a factor of $4$ divides the speed by $\sqrt{4} = 2$, so the outer satellite moves at $\dfrac{v}{2}$ (slower, and it also has a longer path).

The orbit's total energy is $E = -\dfrac{GMm}{2r}$, and at infinity the energy is zero. The binding energy is the amount that must be added: $0 - \left(-\dfrac{GMm}{2r}\right) = \dfrac{GMm}{2r}$.

The burn adds kinetic energy, increasing the total energy (making it less negative), which lifts the satellite to a larger orbit. But orbital speed obeys $v = \sqrt{GM/r}$, so the bigger orbit has a lower speed. The added energy goes mostly into potential energy, and the satellite settles into a slower, higher orbit.

Set the fields equal at distance $x$ from Earth: $\dfrac{GM}{x^2} = \dfrac{G(M/81)}{(d-x)^2}$, so $\dfrac{d-x}{x} = \dfrac{1}{9}$, giving $x = 0.90d$. The null point sits much closer to the smaller mass.

$v_{esc} = \sqrt{\dfrac{2GM}{R}}$, so $\dfrac{v_X}{v_E} = \sqrt{\dfrac{M_X/R_X}{M_E/R_E}} = \sqrt{\dfrac{2}{1/2}} = \sqrt{4} = 2$. Doubling the mass and halving the radius each increase the escape speed, giving $2 \times 1.1\times10^{4} = 2.2\times10^{4}\text{ m/s}$.

$M = \dfrac{4\pi^2 r^3}{GT^2} = \dfrac{4\pi^2 (3.8\times10^{8})^3}{6.67\times10^{-11}(2.36\times10^{6})^2} \approx 6.0\times10^{24}\text{ kg}$, the accepted mass of the Earth. This is how planetary masses are measured from their satellites.

The energy required is the increase in total energy, $E = -\dfrac{GMm}{2r}$: $\Delta E = \dfrac{GMm}{2}\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right) = \dfrac{(4.0\times10^{14})(500)}{2}\left(\dfrac{1}{1.0\times10^{7}} - \dfrac{1}{2.0\times10^{7}}\right) = (1.0\times10^{17})(5.0\times10^{-8}) = 5.0\times10^{9}\text{ J}$.

Speed: $v = \sqrt{GM/r} \propto r^{-1/2}$, so doubling $r$ multiplies $v$ by $\tfrac{1}{\sqrt{2}} \approx 0.71$ (slower). Energy: $E = -\dfrac{GMm}{2r}$, so doubling $r$ halves the magnitude, making $E$ less negative (it increases). The higher orbit is slower but more energetic.